Ta có: \(\left(1-\sqrt[3]{2}\right)^3\)

\(=1^3-3\cdot1^2\cdot\sqrt[3]{2}+3\cdot1\cdot\sqrt[3]{2^2}-\left(\sqrt[3]{2}\right)^3\)

\(=1-3\sqrt[3]{2}+3\cdot\sqrt[3]{4}-2=3\sqrt[3]{4}-3\sqrt[3]{2}-1\)

=>\(\sqrt[3]{3\cdot\sqrt[3]{4}-3\cdot\sqrt[3]{2}-1}=\sqrt[3]{\left(1-\sqrt[3]{2}\right)^3}=1-\sqrt[3]{2}\)

Ta có: \(A=\sqrt[3]{2}+\sqrt{7+2\sqrt{10}}+\sqrt[3]{3\cdot\sqrt[3]{4}-3\cdot\sqrt[3]{2}-1}\)

\(=\sqrt[3]{2}+1-\sqrt[3]{2}+\sqrt[2]{\left(\sqrt5+\sqrt2\right)^2}\)

\(=1+\sqrt5+\sqrt2\)

A=1/2-1/2^3+1/2^5 -1/2^7+......-1/2^33

2²A=\(\frac{1}{2^3}-\frac{1}{2^5}+\frac{1}{2^7}-\frac{1}{2^9}+...-\frac{1}{2^{35}}\)

A+4A=5A=\(\frac{1}{2}-\frac{1}{2^{35}}\)

\(A=\frac{\frac{1}{2}-\frac{1}{2^{35}}}{5}\)

a) Ta có:

\(A=\left(a-4\right)\left(a+5\right)-\left(a-5\right)\left(a+4\right)\)

\(=\left[\left(a-4\right)a+5\left(a-4\right)\right]-\left[\left(a-5\right)a+4\left(a-5\right)\right]\)

\(=\left[a^2-4a+5a-20\right]-\left[a^2-5a+4a-20\right]\)

\(=a^2-4a+5a-20-a^2+5a-4a+20\)

\(=\left(a^2-a^2\right)+\left(-4a+5a+5a-4a\right)+\left(-20+20\right)\)

\(=0+2a+0\)

\(=2a\)

b) Ta có:

\(B=\left(2-a\right)\left(a+7\right)-\left(a-1\right)\left(a+2\right)\)

\(=\left[\left(2-a\right)a+7\left(2-a\right)\right]-\left[\left(a-1\right)a+2\left(a-1\right)\right]\)

\(=\left[2a-a^2+14-7a\right]-\left[a^2-a+2a-2\right]\)

\(=2a-a^2+14-7a-a^2+a-2a+2\)

\(=\left(2a-7a+a-2a\right)-\left(a^2+a^2\right)+\left(14+2\right)\)

\(=-6a-2a^2+16\)

Ta có

B   =   2 a − 3 a + 1 − a − 4 2 − a a + 7   =   2 a 2   +   2 a   –   3 a   –   3   –   ( a 2   –   8 a   +   16 )   –   ( a 2   +   7 a )     =   2 a 2   +   2 a   –   3 a   –   3   –   a 2   +   8 a   –   16   –   a 2   –   7 a     =   -   19

Đáp án cần chọn là: D

đi mk giải cho 100 % luôn ko lừa đâu vì mk học lớp 7 mà đối với mk  bài này dễ lắm

Lời giải:

$A=(2x-1)(4x^2+2x+1)-7(x^3+1)=(2x)^3-1^3-7x^3-7$

$=8x^3-1-7x^3-7=x^3-8$

b.

Tại $x=\frac{-1}{2}$ thì: $A=(\frac{-1}{2})^3-8=\frac{-65}{8}$

\(a,A=8x^3-1-7x^3-7=x^3-9\\ b,x=-\dfrac{1}{2}\Leftrightarrow A=-\dfrac{1}{8}-9=-\dfrac{73}{8}\)